1 solutions

  • 0
    @ 2025-11-5 19:21:29

    C :

    int main(int argc, char* argv[])
    { 
    	double m,n,sum,i;
    	while(~scanf("%lf%lf",&m,&n))
    	{
    		sum=m;
    	  for(i=0;i<n;i++)
    	  {   
    		  sum+=m;
    		  m*=1.0/2;
    	  }
    	  sum-=2*m;
    	 printf("%.2lf %.2lf\n",m,sum);
    	}
    	return 0;
    }
    
    

    C++ :

    #include<iostream>
    #include<cstdio>
    using namespace std;
    int main()
    {
    	double n,sum;
    	int m;
    	cin>>n>>m;
    	sum=-n;
    	for (int i=1; i<=m; i++)
    	{
    		sum+=2*n;
    		n/=2;
    	}
    	printf("%.2lf %.2lf\n",n,sum);
    	return 0;
    }
    
    • 1

    Information

    ID
    19285
    Time
    1000ms
    Memory
    128MiB
    Difficulty
    (None)
    Tags
    (None)
    # Submissions
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